\(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [124]

Optimal result

Integrand size = 26, antiderivative size = 181 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {13 \cot (c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a^2 d} \] Output:

3*I*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+1/4*I*arctanh(1/2* 
(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(3/2)/d+1/3*cot(d*x+c) 
/d/(a+I*a*tan(d*x+c))^(3/2)+13/6*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(1/2)-7 
/2*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.81 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {36 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {3 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{3/2}}-\frac {2 \cot (c+d x) \left (-21+29 i \cot (c+d x)+6 \cot ^2(c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{a^2 (i+\cot (c+d x))^2}}{12 d} \] Input:

Integrate[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(((36*I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/a^(3/2) + ((3*I)*Sqr 
t[2]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/a^(3/2) - (2*C 
ot[c + d*x]*(-21 + (29*I)*Cot[c + d*x] + 6*Cot[c + d*x]^2)*Sqrt[a + I*a*Ta 
n[c + d*x]])/(a^2*(I + Cot[c + d*x])^2))/(12*d)
 

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.654, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4081, 25, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (8 a-5 i a \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (8 a-5 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {8 a-5 i a \tan (c+d x)}{\tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (14 a^2-13 i a^2 \tan (c+d x)\right )dx}{a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (14 a^2-13 i a^2 \tan (c+d x)\right )dx}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (14 a^2-13 i a^2 \tan (c+d x)\right )}{\tan (c+d x)^2}dx}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {3 \left (\frac {\int -\cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (7 \tan (c+d x) a^3+6 i a^3\right )dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (-\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (7 \tan (c+d x) a^3+6 i a^3\right )dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (7 \tan (c+d x) a^3+6 i a^3\right )}{\tan (c+d x)}dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {\frac {3 \left (-\frac {a^3 \int \sqrt {i \tan (c+d x) a+a}dx+6 i a^2 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (-\frac {a^3 \int \sqrt {i \tan (c+d x) a+a}dx+6 i a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {3 \left (-\frac {6 i a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {2 i a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \left (-\frac {6 i a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {6 i a^4 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {12 a^3 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {12 i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

Input:

Int[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

Cot[c + d*x]/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((13*a*Cot[c + d*x])/(d* 
Sqrt[a + I*a*Tan[c + d*x]]) + (3*(-((((-12*I)*a^(7/2)*ArcTanh[Sqrt[a + I*a 
*Tan[c + d*x]]/Sqrt[a]])/d - (I*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c 
 + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/a) - (14*a^2*Cot[c + d*x]*Sqrt[a + I*a*Tan 
[c + d*x]])/d))/(2*a^2))/(6*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] 
 && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

rule 4081

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 

rule 4082

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4083

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( 
A*b + a*B)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A 
*d)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T 
an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - 
 a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (-\frac {5}{4 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {1}{6 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {9}{2}}}\right )}{d}\)	\(137\)
default	\(\frac {2 i a^{3} \left (-\frac {5}{4 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {1}{6 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {9}{2}}}\right )}{d}\)	\(137\)

Input:

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d*a^3*(-5/4/a^4/(a+I*a*tan(d*x+c))^(1/2)-1/6/a^3/(a+I*a*tan(d*x+c))^(3 
/2)+1/a^4*(1/2*I*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+3/2/a^(1/2)*arctanh 
((a+I*a*tan(d*x+c))^(1/2)/a^(1/2)))+1/8/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a 
*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 601 vs. \(2 (138) = 276\).

Time = 0.09 (sec) , antiderivative size = 601, normalized size of antiderivative = 3.32 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/12*(3*sqrt(1/2)*(-I*a^2*d*e^(5*I*d*x + 5*I*c) + I*a^2*d*e^(3*I*d*x + 3* 
I*c))*sqrt(1/(a^3*d^2))*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c 
) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d* 
x + I*c))*e^(-I*d*x - I*c)) + 3*sqrt(1/2)*(I*a^2*d*e^(5*I*d*x + 5*I*c) - I 
*a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a 
^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1 
/(a^3*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 9*(-I*a^2*d*e^(5*I*d* 
x + 5*I*c) + I*a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(16*(3*a^2* 
e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d* 
x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a^2)*e^(-2 
*I*d*x - 2*I*c)) + 9*(I*a^2*d*e^(5*I*d*x + 5*I*c) - I*a^2*d*e^(3*I*d*x + 3 
*I*c))*sqrt(1/(a^3*d^2))*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*sqrt(2)*(a^ 
3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I* 
c) + 1))*sqrt(1/(a^3*d^2)) + a^2)*e^(-2*I*d*x - 2*I*c)) - sqrt(2)*sqrt(a/( 
e^(2*I*d*x + 2*I*c) + 1))*(-28*I*e^(6*I*d*x + 6*I*c) - 13*I*e^(4*I*d*x + 4 
*I*c) + 16*I*e^(2*I*d*x + 2*I*c) + I))/(a^2*d*e^(5*I*d*x + 5*I*c) - a^2*d* 
e^(3*I*d*x + 3*I*c))
 

Sympy [F]

\[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\cot ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cot(d*x+c)**2/(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Integral(cot(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.02 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i \, a {\left (\frac {4 \, {\left (21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 13 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}} + \frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} + \frac {36 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}}{24 \, d} \] Input:

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

-1/24*I*a*(4*(21*(I*a*tan(d*x + c) + a)^2 - 13*(I*a*tan(d*x + c) + a)*a - 
2*a^2)/((I*a*tan(d*x + c) + a)^(5/2)*a^2 - (I*a*tan(d*x + c) + a)^(3/2)*a^ 
3) + 3*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2 
)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2) + 36*log((sqrt(I*a*tan(d* 
x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(5/2))/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,13{}\mathrm {i}}{6\,d}+\frac {a\,1{}\mathrm {i}}{3\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{2\,a\,d}}{a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2}\right )\,\sqrt {-a^3}\,3{}\mathrm {i}}{a^3\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^2}\right )\,\sqrt {-a^3}\,1{}\mathrm {i}}{4\,a^3\,d} \] Input:

int(cot(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

- (((a + a*tan(c + d*x)*1i)*13i)/(6*d) + (a*1i)/(3*d) - ((a + a*tan(c + d* 
x)*1i)^2*7i)/(2*a*d))/(a*(a + a*tan(c + d*x)*1i)^(3/2) - (a + a*tan(c + d* 
x)*1i)^(5/2)) - (atan(((-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/a^2)*(- 
a^3)^(1/2)*3i)/(a^3*d) - (2^(1/2)*atan((2^(1/2)*(-a^3)^(1/2)*(a + a*tan(c 
+ d*x)*1i)^(1/2))/(2*a^2))*(-a^3)^(1/2)*1i)/(4*a^3*d)
 

Reduce [F]

\[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\int \frac {\cot \left (d x +c \right )^{2}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i +\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a} \] Input:

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

int(cot(c + d*x)**2/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i + sqrt(tan(c 
+ d*x)*i + 1)),x)/(sqrt(a)*a)